Law of cosines, Ptolemy's, angle chasing on an isosceles triangle inscribed in a circle From HMMT:

Triangle $\triangle PQR$, with $PQ=PR=5$ and $QR=6$, is inscribed in circle $\omega$. Compute the radius of the circle with center on $QR$ which is tangent to both $\omega$ and $PQ$.

I haven't made much progress. I've set $QS=x$ and $SR=y$ to try for Law of Cosines, since we know $\cos\angle QSR$, but that really hasn't lead anywhere. With Ptolemy's, I 've found that $PS=\displaystyle\frac{5(x+y)}{6}$, but unfortunately $PS$ isn't colinear with anything useful (like the line connecting the centers). I also haven't really been able to use the tangent properties.
Hints beyond what I've done or any useful insights would be greatly appreciated!
 A: 
Given that $|PQ|=|PR|=5,\ |QR|=6$,
the area, the height
and
the circumradius of $\triangle PQR$ are
$S=12$, $|PF|=4$
and
$R_0=\tfrac{25}8$, respectively.
Let
$\angle PQR=\alpha$,
$\angle FOE=\phi$.
Assuming that the center of the circle $O_t\in QR$,
we must have $|DQ_t|=|EQ_t|=r$.
\begin{align}
\sin\alpha&=\frac{|PF|}{|PQ|}
=\frac45
,\\
|OF|&=|PF|-R_0=\tfrac78
\tag{1}\label{1}
.
\end{align}
We have two conditions for $r$, $\phi$:
\begin{align}
|QO_t|+|FO_t|&=\tfrac12\,|QR|
\tag{2}\label{2}
,\\
\frac r{\sin\alpha}
+
(R_0-r)\sin\phi
&=3
\tag{3}\label{3}
,\\
(R_0-r)\cos\phi&=|OF|
\tag{4}\label{4}
.
\end{align}
Excluding $\phi$ from \eqref{3},\eqref{4}
and using known values, we get
\begin{align}
r&=\frac{20}9
.
\end{align}
A: I've done this using coordinate geometry. Let the sides of the triangle be along the sides as shown in the figure.  The angle between $PQ$ and $QR$ is $53°$ (approximately) because of the $(3,4,5)$ triangle formed by the sides $PQ,QB,BP$ where $B$ is midpoint of $QR$. So, the equation line along $PQ$ is $3y=4x$.
Now, let the centre and radius of the unknown circle be $C(h, 0) $ and $r$ respectively. We get two constraints from these,

*

*Perpendicular distance from $C$ to the line $PQ$ is $r$.


*$\displaystyle AC=\frac{25}{8}-r$.
You've two equations and two variable and may proceed now
